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This is not homework.

Example 3) (d) of section P.4, rational exponents in Algebra and Trigonometry:

$$\frac{1}{\sqrt[3]{x^4}} = \frac{1}{x^\frac43} = x^{-4/3}$$

Completely rational. Almost absolutely. However, expected to arrive at the same answer by simplifying the denominator first:

$$\frac{1}{\sqrt[3]{x^{4}}} = \frac{1}{\sqrt[3]{x^{3}}\cdot\sqrt[3]{x}} = \frac{1}{x \sqrt[3]{x}} = \cdots er.. \frac{1}{\sqrt[3]{x^{2}}}\qquad?$$

It seemed like a good idea at the time. Now it just seems made up. Please help. I don't know how it got there instead.

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  • $\begingroup$ $x^1\cdot x^\frac 13=x^{1+\frac13}=x^\frac43$. Where did the squared term come from? $\endgroup$ – abiessu Apr 27 '14 at 20:11
  • $\begingroup$ @abiessu aah.. thank you. In the simplest of places. When you post as an answer, I'll select. $\endgroup$ – user8979 Apr 27 '14 at 20:18
  • $\begingroup$ @abiessu as for the squared term, saw the multiply between the two terms and just went for it. $\endgroup$ – user8979 Apr 27 '14 at 20:24
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$$\frac 1{\sqrt[3]{x^4}}=\frac 1{\sqrt[3]{x^3}\cdot\sqrt[3]{x}}=\frac 1{x^1\cdot x^\frac 13}=\frac 1{x^{1+\frac 13}}=\frac 1{x^{\frac 43}}=x^{-\frac 43}=\left(x^{-\frac 23}\right)^2$$

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$ \LARGE\ \frac{1}{x^\frac{4}{3}} = \frac{1}{x^\frac{3}{3}} * \frac{1}{x^\frac{1}{3}} = \frac{1}{x^1*x^\frac{1}{3}} = \frac{1}{x^{\frac{1}{1}+\frac{1}{3}}} = \frac{1}{x^\frac{4}{3}} = x^\frac{-4}{3} $

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